Powering a cellphone

I have this cell phone we dont use anymore, but I want to keep it because it has the picture of a love one as wallpaper on the little screen. So I wanted to do this: use a photo- resistor as a light-dark sensor and with some other components, have the phone screen turn on at dark so that the picture will glow in the dark and then put a picture frame or something on my night table. Now, the issue was this: the phone does not have the battery or the power cord (and I dont want to get them). So what I thought in doing was to power the phone with a LM317 and set it to 4.5 V which is what the old battery used to give. First I tried using regular 1.5 V batteries (3 of them = 4.5V) to see how it works, and it works fine; but when I use the LM317 the phone does not turn on at all. As soon as the circuit turns the phone on, the voltage drops to about 1.9V (at the LM317 output), but why?????Can some one tell me whats wrong? What am I missing? Is it something about the current that needs to go to the phone? (the phone draws about 200 to 400 mA, the transformer gives about 7.5V output; power on the LM317 would be ~ 1.2W).Thanks.

Can't you transfer the picture to a new cell phone or to the computer?

Also, there is the question of the thermal shutdown, which happens if the load is draining too much power and the regulator overheats. The thermal shutdown prevents the regulator from overheating to a destructive temperature, but can cause the load to have a lower voltage. Is your LM317 burning hot? If so, replace the heatsink by a bigger one. There is also the last possibility of you having a counterfeit LM317 (I heard stories, but never came across with one before). Before further analysis, please could you tell me the colours of your resistors?

By the way, nice drawing. Not the best work of art, but always pleased to see a sketch by hand.

It sounds as though the power available from your 7.5 volt DC source is too low. Is there any information on the power supply that tells what the current available from the 7.5V source happens to be?

I am assuming that the 7.5V is DC and not AC. Am I correct in making this assumption?

hgmjr

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Thanks for your reply. Yes, the transformer is DC and is rated 400mA. I see your point, and so I decided to try with other transformers. I tried with one rated 500mA and the "voltage drop" this time was of 2.5V, ......so it looks like that is the problem (the transformers are just trying to keep the same power; if the device draws lots of mA and the power of it is low, the voltage drops). The higher the Amp rating , the less the voltage drop I would have. I thing if I would get something like 1Amp there would be a voltage drop, but not below 4.5, therefore the phone would work.
Thanks again.

Can't you transfer the picture to a new cell phone or to the computer?

Also, there is the question of the thermal shutdown, which happens if the load is draining too much power and the regulator overheats. The thermal shutdown prevents the regulator from overheating to a destructive temperature, but can cause the load to have a lower voltage. Is your LM317 burning hot? If so, replace the heatsink by a bigger one. There is also the last possibility of you having a counterfeit LM317 (I heard stories, but never came across with one before). Before further analysis, please could you tell me the colours of your resistors?

By the way, nice drawing. Not the best work of art, but always pleased to see a sketch by hand.

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Thank for your reply. There is no way to transfer the picture, all the connectors are damaged. That's way I am trying to do this; in addition to learn the theory behind this issue.
The thermal, probably is not a problem here. The phone only draws about 400mA for 4 or 5 seconds (while all the key leds turn on and off) after the phone is turn on. Then it draws about 230mA. The heat sink doesn't even get a chance to get hot. Unless the 317 sences to much mA's and it does what you are saying.
I think the problem could be the power of the transformer being used. See above answer.
With the colors of the resistors, I don't have the ckt in front of me but i remember the top one is a 270 ohms. The formula for this LM317 I think is
Vo=1.25(1+(R2/R1)).
Thanks again